Questions: Watch your cholesterol: The mean serum cholesterol level for U.S. adults was 198, with a standard deviation of 39 (the units are milligrams per deciliter). A simple random sample of 106 adults is chosen. Use O Cumulative Normal Distribution Table if needed. Round the answers to at least four decimal places. Part 1 of 3 (a) What is the probability that the sample mean cholesterol level is greater than 204? The probability that the sample mean cholesterol level is greater than 204 is.

Solution Steps

The mean serum cholesterol level for U.S. adults is given as \( \mu = 198 \) mg/dL, with a standard deviation of \( \sigma = 39 \) mg/dL. A simple random sample of \( n = 106 \) adults is chosen.

We need to find the probability that the sample mean cholesterol level is greater than \( 204 \) mg/dL. This can be expressed mathematically as: \[ P(\bar{X} > 204) \]

The standard error of the mean (SEM) is calculated using the formula: \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{39}{\sqrt{106}} \approx 3.814 \]

To find the Z-score corresponding to the sample mean of \( 204 \), we use the formula: \[ Z = \frac{\bar{X} - \mu}{\text{SEM}} = \frac{204 - 198}{3.814} \approx 1.5839 \]

Using the Z-score, we can find the probability: \[ P(\bar{X} > 204) = 1 - P(Z < 1.5839) = 1 - \Phi(1.5839) \] From the cumulative normal distribution table, we find: \[ \Phi(1.5839) \approx 0.9434 \] Thus, \[ P(\bar{X} > 204) = 1 - 0.9434 = 0.0566 \]

Final Answer